step 1 the taylor series formula is given. f(a)+f(a)(x - a)+f(a)/2!(x - a)^2 + f(a)/3!(x - a)^3 +…

Explanation:

Step1: Recall derivative rules

For $f(x)=\frac{7}{x}=7x^{- 1}$, using the power - rule for differentiation $\frac{d}{dx}(x^n)=nx^{n - 1}$, we have $f^{\prime}(x)=7\times(-1)x^{-2}=\frac{-7}{x^{2}}$, $f^{\prime\prime}(x)=7\times(-1)\times(-2)x^{-3}=\frac{14}{x^{3}}$, $f^{\prime\prime\prime}(x)=7\times(-1)\times(-2)\times(-3)x^{-4}=\frac{-42}{x^{4}}$, $f^{(4)}(x)=7\times(-1)\times(-2)\times(-3)\times(-4)x^{-5}=\frac{168}{x^{5}}$.

Step2: Evaluate at $a = - 3$

When $a=-3$:

• $f(-3)=\frac{7}{-3}=-\frac{7}{3}$.
• $f^{\prime}(-3)=\frac{-7}{(-3)^{2}}=-\frac{7}{9}$.
• $f^{\prime\prime}(-3)=\frac{14}{(-3)^{3}}=-\frac{14}{27}$.
• $f^{\prime\prime\prime}(-3)=\frac{-42}{(-3)^{4}}=-\frac{42}{81}=-\frac{14}{27}$.
• $f^{(4)}(-3)=\frac{168}{(-3)^{5}}=-\frac{168}{243}=-\frac{56}{81}$.

The values in step 2 for $f^{\prime\prime}(-3)$ and $f^{(4)}(-3)$ in the original were incorrect. The correct values for the Taylor - series coefficients are based on the above evaluations.

Answer:

$f(-3)=-\frac{7}{3}$, $f^{\prime}(-3)=-\frac{7}{9}$, $f^{\prime\prime}(-3)=-\frac{14}{27}$, $f^{\prime\prime\prime}(-3)=-\frac{14}{27}$, $f^{(4)}(-3)=-\frac{56}{81}$